If in a Delta ABC,sin^3 A + sin^3 B + sin^3 C | vedclass

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(D) We use the algebraic identity: $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$.Substituting $x = \sin A$,$y =

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$\Delta ABC$ is an equilateral triangle

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(D) We use the algebraic identity: $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$.Substituting $x = \sin A$,$y = \sin B$,and $z = \sin C$,we get:$\sin^3 A + \sin^3 B + \sin^3 C - 3 \sin A \sin B \sin C = \frac{1}{2}(\sin A + \sin B + \sin C)[(\sin A - \sin B)^2 + (\sin B - \sin C)^2 + (\sin C - \sin A)^2]$.Since the given equation is $\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$,the expression equals $0$.This implies either $\sin A + \sin B + \sin C = 0$ (which is impossible for a triangle as $A, B, C > 0$ and $A+B+C = \pi$) or $(\sin A - \sin B)^2 + (\sin B - \sin C)^2 + (\sin C - \sin A)^2 = 0$.This occurs only when $\sin A = \sin B = \sin C$,which implies $A = B = C = 60^\circ$.Thus,$\Delta ABC$ is an equilateral triangle.

If in a $\Delta ABC$,$\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \cdot \sin B \cdot \sin C$,then:

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